C语言中的char指针初始化 [英] char pointer initialization in C

问题描述

我不清楚字符指针及其工作原理.

I am not so clear on character pointer and how they work.

该程序已构建，但在运行时崩溃.

The program builds, but crashes when I run it.

 char *ab = NULL;
 //ab = "abc123"; // works fine
 sprintf(ab, "abc%d", 123); // this line seems to crash the program

当sprintf将(char * str)作为第一个参数时，我不知道这怎么错.

I don't get how this can be wrong, when sprintf takes in a (char * str) as a first argument.

任何人都可以向我解释一下吗?

Can anyone please explain this to me?

推荐答案

您尚未分配要与ab一起使用的内存.

You have allocated no memory to use with ab.

第一个分配起作用是因为您要向ab分配一个字符串常量:"abc123".常量字符串的内存由编译器代表您提供:您不需要分配此内存.

The first assignment works because you are assigning to ab a string constant: "abc123". Memory for constant strings are provided by the compiler on your behalf: you don't need to allocate this memory.

例如，在将ab与sprintf，您需要使用malloc分配一些内存，并将该空间分配给ab:

Before you can use ab with e.g. sprintf, you'll need to allocate some memory using malloc, and assign that space to ab:

ab = malloc(sizeof(char) * (NUM_CHARS + 1));

然后，只要您使用malloc留出了足够的空间，您的sprintf就可以工作.注意:+ 1用于为空终止符.

Then your sprintf will work so long as you've made enough space using malloc. Note: the + 1 is for the null terminator.

或者，您可以通过将其声明为数组来为ab腾出一些内存:

Alternately you can make some memory for ab by declaring it as an array:

char ab[NUM_CHARS + 1];

在不为ab分配内存的情况下，sprintf调用将尝试写入NULL，这是未定义的行为.这是导致崩溃的原因.

Without allocating memory somehow for ab, the sprintf call will try to write to NULL, which is undefined behavior; this is the cause of your crash.

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